标签：define 部分 second str 转移 space tac end SQ
题意：给定一个表达式，然后让你添加 n 个加号，m 个减号，使得表达式的值最大。
析：首先先要建立一个表达式树，这个应该很好建立，就不说了，dp[u][i][0] 表示 u 这个部分表达式，添加 i 个符号，使值最大，dp[u][i][1] 表示 u 个部分表达式，添加 i 个符号，使用值最小，这里添加的符号可能是加号，也可以是减号，就是最小的那个，因为题目说了min(n, m) <= 100，一开始我没看到，就 WA8 了。然后在状转移的时候，分成两类，一类是添加的是加号，那么 dp[u][i][0] = max{dp[lson][j][0] + dp[rson][i-j][0], dp[lson][j][0] - dp[rson][i-j-1][1] }，同理可以得到添加减号的时候。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e4 + 20;
const int maxm = 1e6 + 10;
const LL mod = 1000000000000000LL;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){ int x; scanf("%d", &x); return x; }
int ch[maxn][2];
int num[maxn];
string s;
int rt, x;
void dfs(int l, int r, int &rt){
if(l == r){
rt = l;
ms(ch[rt], -1);
return ;
}
int det = 0;
for(int i = l; i <= r; ++i)
if(s[i] == ‘(‘) ++det;
else if(s[i] == ‘)‘) --det;
else if(s[i] == ‘?‘ && det == 1){
rt = i;
dfs(l+1, i-1, ch[i][0]);
dfs(i+1, r-1, ch[i][1]);
num[rt] = num[ch[i][0]] + num[ch[i][1]] + 1;
return ;
}
}
int dp[maxn][105][2];
void dfs(int u){
if(ch[u][0] == -1){
dp[u][0][0] = dp[u][0][1] = s[u] - ‘0‘;
return ;
}
int tt = min(num[u], x);
int ls = ch[u][0], rs = ch[u][1];
int t = min(tt, num[ls]);
dfs(ls); dfs(rs);
for(int i = 0; i <= tt; ++i){
int &mmax = dp[u][i][0]; mmax = -INF;
int &mmin = dp[u][i][1]; mmin = INF;
for(int j = 0; j <= t; ++j){
if(i - j > 0 && i - j - 1 <= min(x, num[rs])) mmax = max(mmax, dp[ls][j][0] + (n <= m ? dp[rs][i-j-1][0] : -dp[rs][i-j-1][1]));
if(i - j >= 0 && i - j <= min(num[rs], x)) mmax = max(mmax, dp[ls][j][0] - (n <= m ? dp[rs][i-j][1] : -dp[rs][i-j][0]));
if(i - j > 0 && i - j - 1 <= min(x, num[rs])) mmin = min(mmin, dp[ls][j][1] + (n <= m ? dp[rs][i-j-1][1] : -dp[rs][i-j-1][0]));
if(i - j >= 0 && i - j <= min(num[rs], x)) mmin = min(mmin, dp[ls][j][1] - (n <= m ? dp[rs][i-j][0] : -dp[rs][i-j][1]));
}
}
}
int main(){
cin >> s;
cin >> n >> m;
x = min(m, n);
dfs(0, s.sz-1, rt);
dfs(rt);
cout << dp[rt][x][0] << endl;
return 0;
}
CodeForces 935E Fafa and Ancient Mathematics (树形DP)
标签：define 部分 second str 转移 space tac end SQ
原文地址：https://www.cnblogs.com/dwtfukgv/p/9078281.html
